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What is the downwash speed on a typical helicopter?
On a typical 2 or 4 seat helicopter like a Robinson R22/R44, at full throttle at 20ft ASL, if you put a Wind Speed Meter/Anemometer about 10ft below the blades and halfway along a rotor blade's span, about what speed should it read?

What you are searching for is called the induced velocity $v_i$. It can be calculated in a good approximation via the following formula:
$v_i = \sqrt{\frac{F_n}{2 \rho S}}$
With $F_n$ being the weight of the aircraft, $\rho$ the air density, and $S$ the disk area. This formula assumes an "ideal" rotor, therefore air is uniformly accelerated over the rotor plane without losses.
Here are some example values for light, medium and heavy weight helicopters as well as a tilt rotor at sea level for the MTOW (Maximum Take-off Weight):

Robinson R-22 (light weight helicopter): $7.34 \frac{m}{s} \approx 24.1 \frac{ft}{s}$
Airbus H135 (medium weight helicopter): $10.01 \frac{m}{s} \approx 32.8 \frac{ft}{s}$
Leonard AW139 (heavy weight helicopter): $13.55 \frac{m}{s} \approx 44.5 \frac{ft}{s}$
Boeing V-22 Osprey (tilt-rotor helicopter): $20.24 \frac{m}{s} \approx 66,4 \frac{ft}{s}$

Real-life adaptation
In reality however, the rotor plane is not a continuous plane which accelerates air, as assumed by the momentum theory. In reality the airspeed over the rotor plane rises linearly outwards, as the velocity rises linearly over their span (a rotor segment spins faster the more distant it is from the rotor hub). The situation is roughly as follows:

However as you ask about about a point halfway along a rotor blade, the induced velocity is pretty much the mean induced velocity as calculated above. Additionally, the stream tube does not contract very fast, therefore the induce velocity might be a bit higher, but not by much. Sophit states in his answer that the stream tube contracts by a factor of 2 over the rotor diameter, which would be 25 feet for the R22, therefore at 10 feet it is perhaps 10ft/25ft = 0.4 about 40% faster. Given that, an the fact that I estimate the full throttle thrust to be about 20% of hover thrust at MTOW, the induced velocity of an R22 would then be about $11.36 \frac{m}{s} \approx 37,3 \frac{feet}{s}$.

An R22 7.34m/s? Unless I'm misunderstanding something those numbers just don't make any sense to me(that is too slow). Also where do you take into account the 10ft distance bellow the blades on that formula?

@Gabe The 10 ft can only be taken into consideration with a full  computational fluid dynamics solution. U_flow's answer is for a spherical cow in a vacuum (see the words 'good approximation' in the opening), which gets you most of the way there, and certainly allows comparison between disk loadings.

@Gabe no the number is definitely correct. The induced velocity as calculated above stems from momentum theory, therefore it is a mean value across the rotor disk. In reality, the induced velocity rises from the hub to the tip linearly until shortly before the tip, at which point it drops of. However at the location which you pointed at, it hits exactly the mean value. Therefore, the answer is correct. I was also a bit taken back, because by intuition this number should be higher, but helicopters are built to reduce induced velocity to increase efficency.

@Gabe what happens if the induced velocity is too high, is nicely illustrated by the V-22's downwash as seen here: https://www.youtube.com/watch?v=ObfdLy-QlsU. Also compare this answer: https://aviation.stackexchange.com/questions/100669/does-the-osprey-have-a-worse-downwash-than-a-helicopter-of-equal-weight.

I think this only applies for horizontal flight at reasonable speed, when the rotor is constantly moving into fresh still air. In hover the air velocity will be higher as the air starts to spin around the helicopter. Here is a CFD simulation for UH-60; the formula in this answer gives 44 ft/s, which seems to agree quite well.

@jpa that explanation makes more sense and the number makes much more sense also.

@Gabe But 44 ft/s is 13,4 m/s, therefore exactly the value I have given above for a heavy helicopter. Why does that value make more sense?
The equations of the momentum theory given above are for hover flight, not forward flight.

ooops I read 44 m/s. my bad. See @sophit's answer though, I think is a more accurate answer and number.

@Gabe I tried to update my answer to include more real-life effects. However 7.34 m/s is accurate for the mean velocity. I get that this seems low, but this is just how it is! Helicopters are built to accelerate large cross sections of air to relatively low speed, because that is efficent hence their big rotor! The inverse would be things like the harrier jet, which accelerates a small cross-section of air to a very high velocity (due to their jet-engine) but is extremly inefficient, and therefore can only hover for a very small amount of time.

@Uflow excellent work with Sophit, and also there will be (in hover) a vortex velocity component.  I watched some ocean helicopter rescue videos to study the wind flow pattern.  Also interesting is the momentum of the side air being drawn toward the vortex helps tighten it, which will maintain (or even increase) its angular velocity.

@Gabe Be that as it may, but could you accept either sophits or my answer so that this question is finished? It would help the site tremendously.

@U_flow don't get me wrong, I really appreciate the time you took to post your answer, however I really feel that speed would be higher if you put a Wind Speed Meter.  I understand perfectly your point about the large cross section area compared to like a Harrier jet. However that number looks to me something more like those big fans used in Industrial warehouses compared to the rpm of a helicopter(yes I know I'm exaggerating a little bit). But anyway, I'm waiting to see if maybe someone else posts another answer which might have another number.

@Gabe hehe, that is your good right, but: "the good thing about engineering is that you don't have to believe it for it to be right" ;)

@U_flow agree.  Thanks for your passion and understanding. Let's wait a bit more and see if someone else drops something else.

@Gabe Some time has passed. For the sake of closing this question, would you consider accepting one of the two answers?

Airflow around rotors can be extremely complicated (NASA source):

Estimating the airspeed in exactly one point is therefore very difficult and is normally done only if the point belongs to the helicopter. Anyway a good approximation of the average flow field under a rotor can be obtained via the simple momentum theory by which the rotor is assumed as a thin disk imparting the air a jump in its pressure (source):

As visible in this picture, airspeed increase from 0 far above the rotor to $v$ by the rotor and to $w$ beneath it. This particular airspeed shape is termed "vena contracta" (dashed line). For a rotor of disk area $A$ generating a thrust $T$, the momentum theory returns a speed $v$ by the rotor of:
$v=\sqrt{\frac{T}{2 \rho A}}$
As said, under the rotor the airflow keeps on contracting and the speed increasing, reaching a value $w=2v$ at some one-diameter distance under it. Further below airspeed will eventually decrease due to viscosity.
Being in hover thrust equal weight, from that equation it is easy to calculate the speed at the point you indicated in your question. Considering the point halfway between $v$ and $w$ and given the main rotor area of 46m² and the MTOW of 620kg for a R22, we get at sea level:
$w=1.5\sqrt{\frac{620 \times 9.81}{2 \times 1.225 \times 46}}=12m/s$

Good answer, but the air density at sea level is 1.225 kg/m^3. However I did not know that the stream tube contracts so fast! However, please take into account that the point OP is asking about is not 1 Rotor diameter below the rotor plane, but only 10 feet, therefore only 40%.

@U_flow: right thanks, I'll correct the value for the density. One rotor diameter is obviously just a rule of thumb, let's say that a value halfway is maybe more correct.

The instantaneous turbulent flow field is indeed extremely complicated but in these kinds of question and problems we usually care about the values averaged in the Reynolds averaging sense and then the average velocity field is much smoother and the W value at a given point shall be well defined.

authors: Vladimir F Героям слава


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Can foreign hours (solo/sim) be counted?
I am starting flight training overseas first (with sim and some in-aircraft hours) and was wondering should the instructor(local governing body approved instructor but not FAA certified), sign off my solo and I accumulate solo time overseas (+ XC), can these hours flown be counted towards solo (and solo XC) requirements towards PPL and ultimately IFR (the 50 hrs PIC requirement).
And to add on, can an overseas FAA approved FTD used by me and said instructor be counted towards sim time for ppl/IR rating?
since The rules on logging as per 14 CFR 61.51.
Logging training time in a simulator is allowed by 61.51(h):
(h) Logging training time. (1) A person may log training time when that person receives training from an authorized instructor in an aircraft, full flight simulator, flight training device, or aviation training device.
and
a maximum of 2.5 hours of training in a full flight simulator or flight training device representing the category, class, and type, if applicable, of aircraft appropriate to the rating sought, may be credited toward the flight training time required by this section, if received from an authorized instructor.
Must the authorized instructor, in this case, be a FAA certified CFI or does a local one count?

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Why might a flight in turbulence turn off cabin lights and heating?
I was on a flight from Sofia to Bologna in mid-September of 2017. It was an early evening flight (can't remember what time we took off but it was already dark), and soon after takeoff, around the time we reached cruising altitude, something like turbulence started but it lasted throughout the rest of the flight, for over an hour straight.
The turbulence was pretty violent: people were throwing up and the captain announced they were going to lower the altitude (I remember it was significant but I don't remember how much exactly).
I remember it was very cold in the plane throughout the whole flight. It was also pitch dark: they turned off all the lights, except for blue lights in the lower part of the aisle. We were relieved to have landed eventually as it was one hell of a flight.
I fly this particular route very frequently and I've never experienced anything similar. Is it possible that it was just regular turbulence or was something wrong?
I'm not terrified of flying or anything, just super curious about that particular flight. It was a Wizzair flight, the plane was Airbus but I don't remember which model.

Please [edit] to make the question in the title match the question in the body. The title question is "turbulence causing cabin lights/heat to go out" and the body question is "was this normal turbulence".

@FreeMan FWIW, the title was edited by non-OP. The original title was inline with the question body, although perhaps too generic.

Sounds like "one of those nights" when the state of the atmosphere & frontal boundaries just gives you a bunch of turbulence for the entire path of sky that your flight is using. Some days are like that -- it isn't common, but it happens.
Making the cabin cold is a known technique to help with airsickness. Warmer cabins make airsickness more pronounced, and colder cabins do the opposite. It isn't a perfect solution obviously, but it helps, and is definitely in the "you use the tools you have" category.  As the saying goes, "popsicles don't puke"!
Haven't heard of darkening the cabin as a similar technique; it may be been a related thought, or perhaps just their procedures that when service is done and it's nighttime, turn the lights down so that those who want to sleep, can.  Not that many people were likely able to do much of that in this case, from what it sounds like.

Far-fetched idea: lowering the cabin lighting allows passengers' eyes to adapt and maybe at least the people with window seats can see some orientation points on the horizon. From my own experience, being able to visually "anchor" yourself to the Earth helps me with motion sickness.

Isn’t turning the lights down part of some procedure, so that, in case of emergency, passengers’ sight will already be adjusted for the night? Same as for take off and landing – https://aviation.stackexchange.com/questions/8471/why-are-the-lights-inside-commercial-airplanes-turned-off-during-take-off-and-la

@DidierL This question is asking about the enroute portion of the flight; that question only validates a common practice of lowering the lights for dark-adaption at takeoff & landing. What JörgWMittag suggested may be what was the rationale here, but it's not a standard procedure.

Thanks for the insights, guys, much appreciated. I keep coming back to that particular flight because it was unlike anything I had experienced before (or since, for that matter) and I was curious if it sounds like a normal although less common situation or we might have been in some real trouble there (and if so, what kind). What you've commented makes sense (thanks for whoever moderated and updated the title, wasn't sure what to go with initially).

@Dixieland Certainly not "real trouble" in the sense that the aircraft itself would be in danger. The airframe can handle much, much more turbulence than the fragile human cargo inside. Severe turbulence can be a safety issue in that somebody can get quite some bruises or even break an arm if they aren't properly strapped to their seat, but that's still far from snapping the wings off or so. If the flight was in real trouble, it would have diverted to a nearby airport instead of proceeding to your planned destination.

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Can someone explain the damage distribution on this aircraft that flew through a hailstorm?

An A320 flew into a hailstorm 20mins from landing. I’m intrigued by the damage distribution. Seems like everywhere else wasn't touched at all.  Why?

I'd love to see the engines of that plane

@Jpe61 Jet engines eat hail for lunch.

@reirab more like it is a requirement that they eat hail for lunch, and are specifically tested against them to make sure they do not fail. Engines that don't pass are not certified and cannot be used.

@Nelson Yep, that link is to the CFRs that specify the testing requirements for exactly that. :)

Props to that windshield tho

@AlanB like with the jet engines being able to suck up hail without stopping work, the windshield is designed to not shatter under heavy bombardment for the same reasons.  If it was smashed in the pilots would be getting hammered instead which would have far more serious consequences than just bad forward visibility while landing.

I posted this as a comment below, but there is considerable denting of the aluminum as well - the paint must adhere better to it than it does to fiberglass.  If you look along the right side of the aircraft as we see it at the line where the reflection changes from skybridge to open sky, you can see a number of dents from hail.  The same is true of the left side reflections, many of which should be straight lines but instead are all squiggly.  Presumably the entire forward facing aluminum surface is similarly dented, but the lighting in this image doesn't show the damage.

Fun fact: the affected plane is from the Austria Airlines, colloquially known as AUA. "Aua!" is also a colloquial exclamations in German indicating pain, similar to "ouch!" in English. So, when the pilots flew the AUA plane through the hailstorm, the AUA plane said: Aua!

Although the exterior of a jet airliner may just look like uniform aluminum with windows, the construction is actually much different than the outward appearance suggests.
The front of modern jet airliners are what's called a radome, which is a portmanteau of "radar" and "dome".  The radome houses a radar antenna, and the material covering the radome is much more fragile than the rest of exterior surface of the aircraft.  This is to improve the performance of the antenna.  Behind the radome is a structural wall that limits any impact to the radome's exterior from affecting the body of the aircraft.
This is why you see the exterior of the radome in such bad condition, but the rest of the aircraft's exterior in relatively such good condition.
Regarding the windshields, the forward windshield will almost always be exposed to the majority of weather-related impacts.  This is because the aircraft is flying in a forward direction (hopefully!), thus causing the greatest forces on those windows.  A simple way to think about this is to think about an automobile.  The front windshield of an automobile will be most affected by driving through hail, as opposed to any side windows.  When a vehicle (or aircraft) is not moving, however, impacts (and damage) will largely be determined by the installed angles of the windows.

@Bianfable Thank you for the edit... sorry it got deleted while I was editing at the same time, but I restored your improvement.  I'm done editing now, so feel free to make any other edits you like.

"to improve the performance of the antenna" is the understatement of the year. The radar wouldn't work at all if it was covered by metal (such as the aluminium skin used on the rest of the airframe), since metals are opaque to RF radiation. That's why the radome simply has to be plastic (fibreglass).

Do you have any idea why port portion of radome and port window got more damage than starboard side?

@user721108 best theory for biased damage is that the hail was falling at an agle due to winds higher up. So the plane was most likely flying straight, but the hail carried extra sideways velocity.

@TooTea But it is possible to make a fiberglass construction that would survive the hailstorm.. it would just be thick, and thus attenuate signals.

It would appear to me that once the radome had disintegrated, there would be violent radial airflow that would tend to carry a substantial proportion of the impactors away from the fuselage. That also applies to e.g. the leading edge of the wings: they have a strongly divergent radial airflow at all times.

@jpe61 So the hail was moving with the air, but the plane was side-slipping?

@DJohnM no the plane flies straight in its own mass of air. Hail on the other hand gained horizontal velocity before hitting the plane, when falling through an higher altitude layer of air moving sideways relative to the layer in which the plane flies. Hail kept some of that sideways velocity due to inertia, therefore hit the plane at an angle.

@DJohnM I suspect the other way around, but either one is possible. Planes are usually flown "dead straight" without sideslip for efficiency, except for short final in some cases, and flare of course. Conditions that produce hail of significant size usually also produce heavy windshear (thunderstorms), so the hail may have travelled a down in an air mass moving in direction x, and then fell into an airmass moving in direction y, and hit the airplane with residual horizontal speed from previous airmass.

Please note that the hail did not hit the aircraft, it was the other way around: plane flew into the hail at somewhere between 200 and 500 mph, hail falling at less than 100

@MikeB it's all about your reference frame. Einstein: "Does Oxford stop at this train?", or XKCD

@chrisH  No, not really. The point I was making is that the hail did not fall down across the plane to any meaningful extent, rather the plane flew horizontally through the hail, hence the strike pattern being very different from what people expect.

@MikeB that's exactly what reference frames mean in physics.  Motion is all relative. You're thinking in a ground reference frame, or that of the air.  The vectors add up to the same impact angle anyway, hence if looking out of the cockpit windscreen you'd see the hail coming directly towards you, with (little to) no sideways or vertical motion.

@ChrishH  I understand reference frames. But they are not relevant here. Many of the comments on this page talk about the hail 'falling' relative to the plane, which is basically incorrect, eg Jpe61

The windscreen.
As you can see from the photo, hailstones struck the forward windows almost perpendicularly, and the side windows almost parallel. The window material is layers of glass and plastic, and glass is hard and brittle, so impacts up to a certain force threshold will show no damage at all, then above that will cause cracking and cratering.
That's why the forward windows are beaten opaque while the side windows aren't even scratched. This is not surprising.
The port windscreen (right side in the photo) looks to be seconds from collapse. Even a small amount of glass striking the pilot in the face at 300Kts would be a serious distraction. I can only wonder how close this flight was to total loss.
The radome.
I don't know why the fiberglass radome took such damage while the surrounding aluminum seems totally unscathed. The visible damage seems to change suddenly at the boundary while AOA does not. Perhaps there are lots of little dents that don't show in the photo.
I am guessing that aluminum, being elastic and tough, dents upon impact and then springs back into shape. Fiberglass -- glass fibers embedded in resin -- being stiff and brittle, cracks upon similar impact. So aluminum recovers between multiple strikes (until metal fatigue sets in) while fiberglass accumulates damage.
Obviously the center of the radome, perpendicular to the hailstone attack, accumulated so much damage that it collapsed inward, allowing the pressure inside the antenna compartment to instantly rise to a titanic level and pop the radome like a party balloon. That's why most of the radome is missing instead of hanging there in shreds.
Then the hailstorm quickly battered the antenna itself into useless junk, concealing all its brother hailstorms from further detection by this particular aircraft.

LOL, "concealing all its brother hailstorms from further detection"

A couple of glass shards at 300kts might be distracting, but I think a bunch of never ending ice balls are even worse.

I strongly doubt the windows were close to collapsing inwards. It's much easier to prevent them from falling apart than building them in a way so they stay transparent on impacts.

I second the implications that the elasticity of the aluminum helped limit damage to it. Also, the point about the windows kind of applies to the rest of the skin of the aircraft as well, most of the impacts on the wings and fuselage would have been at comparatively shallow angles as well, so the amount of damage they could cause is more limited than any impacts on the radome. I would expect the leading edges of the wings to be a bit more beat-up than the rest of the plane for the same reason.

Composites can be quite brittle, so the hail would erode the surface (those dark patches? They reduce with the impact angle as you'd expect) and weaken the structure. Then once the radome started to crack, the airflow would try to tear it.

Hypothetically, with modern instruments and landing aids functioning, the crew could have landed the plane on a runway with zero external visibility, correct?

Aircraft aluminum is thicker than you might think. It most likely simply didn't dent at all. Remember that this stuff is designed to take an impact from a several-pound bird at 200-300 mph without noticeable damage and, more importantly, is designed to withstand the roughly 7.5 pounds of force per square inch (~0.5 kg/cm^2) attempting to blow the plane apart from the inside at any given time during cruise. The bag tags I have that are cut out of an airliner fuselage are about 1/8" (3 mm) thick. The ability to shrug off hail without damage is a design requirement.

Airplane windscreens are multilayer constructions.  The outermost layer is certainly heavily damaged, but the inner layers are still up to the task of keeping the wind out and the pressure in.

"allowing the pressure inside the antenna compartment to instantly rise to a titanic level and pop the radome like a party balloon", can you explain why this would be the case?
The radome is unpressurized, so I don't see what should cause a big pressure difference.
Also, there are plenty of cases where the radome just gets dented (without popping like a party balloon)

@Criggie that plane was in fact landed without complications: https://noe.orf.at/stories/3260642/ IDK how much visibility the pilots had left.

@ROIMaison the radome normally doesn't have any significant air pressure inside it. However, once it's cracked open, 2-300 MPH wind gets inside and, having nowhere else to go, builds pressure on the inside, which the radome isn't designed to withstand. F1 car bodies are built out of lightweight carbon fiber and can withstand 200MPH wind just fine. However, a little crack or chip will let the air in, and, with no place else to go, will rip bodywork right off without a second thought.

@Criggie BA Flight 009 (https://en.wikipedia.org/wiki/British_Airways_Flight_009) landed with an opaque windscreen and only lateral positioning guidance

@reirab If you look at the right side of the image, there is considerable denting of the aluminum visible.  You can see it along the line where the reflection switches from skybridge to sky.  I suspect that this picture wasn't taken to accentuate the damage to the hull but instead show the devastation of the radome and windscreen.  The same is true of the reflections on the image left, but less clearly.  If nothing else, these reflections should have lots of straight lines but they're all squiggly because of the dents.

@Jason Patterson: Well it's inconclusive but I think you may be right. If the aluminum really is dented, then the visible damage boundary may simply due to the fact that paint is easily chipped from the fiberglass but sticks to the deformed metal.

Oh, good catch @JasonPatterson! If you click to embiggen the image, you can see loads of wrinkles in the AL on the left side (as we see it) fanning out along the AL from where the remains of the radome attach (about 9 o'clock).

Just to add something to the other answers:
What it might have happened to the radome is that it has been damaged enough to loose its structural integrity and the missing piece(s) has just flown away.
Regarding the other parts of the aircraft, it is indeed very likely that also the fairing of the wing-fuselage attachment and the engine  cowlings got damaged like on this B777 but unfortunately without better/other pictures of the incident it's impossible to confirm that:

Note in this picture how the leading edge of the inlet has been also damaged: the leading edge of the wing and the rest of the nose might have suffered the same fate as well.

And look at how the angle changes to the right and left of the radome.  The hail would slip off at this point, and not go smashing in at 90º.  Other answers have kind of said this, but I think that's the primary difference at this particular part of the aircraft.

Good find with those photos.  Does the source indicate if there is any structural damage there, or primarily cosmetic (and, of course, reduced performance)?


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Who determines the minimum area needed for a sky diving dropzone?
Who or what determines the minimum area required for a sky diving dropzone at an airport?

Depends on the dropzone.
If the dropzone is a member of the United States Parachute Association, it must meet the following Basic Safety Requirements (2024 SIM page 7) for the landing area:

Areas used for skydiving should be unobstructed, with the following minimum radial distances to the nearest hazard: [S]

solo students and A-license holders—330 feet
B- and C-license holders and all
tandem skydives—165 feet
D-license holders—40 feet

Hazards are defined as telephone and power lines, towers, buildings, bodies of water, highways, vehicles, and trees covering more than 32,292 square feet. However, clusters of trees must not be greater than 10% of the designated landing area.

It is up to the dropzone what the minimum license required is to jump there. So, if the dropzone requires a D license to jump there, then the minimum landing area size is a 40 foot radius circle.
Besides that, the minimum area depends on the number of staff and jumpers and their comfort/ business requirements, and the type and number of airplanes.
Technically speaking, the airplane tie downs, the runway, and the parking lot could be part of the general airport and not dedicated to the dropzone.
And to really push it, you could handle paperwork, payment, and packing parachutes right in the landing area between loads.
So the strict minimum is just a big enough landing area to safely land a parachute.


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Why does one airliner fly along the coast and the other doesn't?
Found this odd to see the same airliner, same origin, same destination, and same time, yet different paths. More interesting is seeing that one aircraft is flying along the coast while the other is flying straight trough the pacific. Why is this, what would be the reasoning behind to take different paths?

Zoom In of AAL flight to show that there is actual geo data(purple) at this point(no dotted lines)

Screenshot taken on 2024-04-08

Looking at the ANA105 flight history shows they regularly switch between the two paths, with additional variations. More often the "straight west" route the last few days, though.

Related: Why are westbound transatlantic routes located hundreds of km away from eastbound routes?

The routes for the flights shown in the question are correct (see the routes at the end of this answer). Airlines will typically try to choose the fastest route for their flights. Time is money. One of the most common reasons for picking a longer route is weather.

You can see from the current winds aloft that the more northerly great circle route goes through an area of high headwinds of the west coast of the US (shown in the very light yellow color). At around the same time after departure, the route that heads straight west goes through an area of very low wind instead (the area that goes to green and then darker blue).
The AAL flight left 22 minutes later than the ANA flight, but is only currently forecast to get in 29 minutes later. This means the AAL flight will only be about 7 minutes longer, despite having a route that's over 200 miles longer.
These flights are almost 11 hours long, so the routes have to be chosen based on weather forecasts many hours in the future. Small deviations can be made enroute but a large choice such as between these two routes has to be made early. It's always a roll of the dice what the weather will actually do. Maybe the weather didn't turn out quite how AAL hoped, but 7 minutes is pretty small over a flight almost 11 hours long.
For reference, here are the two filed routes shown on FlightAware. You can see the latitude/longitude pairs on the ANA flight are going up above 50 degrees north, while the AAL flight's waypoints are down near 30 degrees north.
ANA105:
SUMMR2 MCKEY LIBBO BRINY BOARS AMAKR REDWD UNVER 4200N/13000W 4600N/14000W 5000N/15000W 5200N/16000W 5300N/17000W 5200N/18000E CHIPT CARTO CUTEE KALIG KALNA OTR5 ADNAP Y807 LALID Y804 INUBO Y108 MESSE
AAL169:
SUMMR2 DINTY DUETS BARAZ 3400N/14000W 3400N/15000W 3400N/16000W 3400N/17000W 3300N/18000E 3300N/17000E 3400N/16000E 3500N/15000E SEALS OTR13 LAPIL Y811 NOGIX Y807 LALID Y804 INUBO Y108 MESSE MESSEN

20 ° difference in latitude!!!!  had no idea

@Fattie cos its such a long distance.  One could imagine two flight paths exactly halfway around the globe, 90 or 180 degrees apart and they'd still have nominally the same distance travelled.

@Criggie sure, I understand the math of great circle metrics - it['s just surprising eh!!!

"Airlines will typically try to choose the fastest route for their flights." I think you'll find a more accurate answer will be that airlines pay for flight planning software to calculate the most efficient flight plan - which using forecast winds along with anticipated aircraft performance (booked passengers, aircraft technical information) - will run various models to find the most efficient routing and cruise alt. My company uses Lido, flight plans are typically only available ~5 hours before - so they are using the most up to date PAX, weather and aircraft data in the modelling.

To add on to @fooot's answer,
I put the 2 routings into SimBrief (https://dispatch.simbrief.com/options/new)  and used the "Calculate & Compare" tool which gives me the ability to see the difference in flight times, fuel burnt, wind component etc. between 2 generated flight plans. Now the 2 aircraft types are not exactly the same, but for comparison purposes I used the B787-9 with GE engines and a cost index of 100, and kept the payload values the same for both flights. SimBrief calculated that the AAL flight would travel an extra 307 nautical miles, and burn about 400 or so kilos of extra fuel. This isn't a lot, especially for such a long flight, but the AAL flight would also have an extra 12 minutes of flight time.
There could also be different flight planning and dispatch philosophies between AAL and ANA, with the ANA guys staying a bit lower and cruising a bit faster to negate some of the effects of the strong headwinds along their path, in return burning more fuel.
Additionally, as the airspace over the Pacific is not under radar control, it could be possible that the AAL flight couldn't secure a slot on the tracks it wanted to take because the ANA flight departed first. Not sure of the separation standards over the Pacific, but the 2 flights pushed back 5 minutes apart and took off 13 minutes apart at LAX, if they were to fly the same route AAL's pushback would certainly be delayed by at least a few minutes (10 minutes separation minimum I believe is what they use, have a look at PBCS). Sometimes ATC will offer a different route so you won't have to take on a delay, or ask you to take on a different cruising level. All comes down to a decision by the flight crew and dispatch.

The two dispatch offices may be using different wind forecast products, or may have planned the routes at different times; forecasts can change in the hours before a flight, and a fairly small shift in forecast winds might make a route that used to look optimal no longer look as efficient under an updated forecast.  Also, depending on flight-specific factors (aircraft type, engines, gross weights), they may be planning different cruise altitudes & step-climb profiles, which could also change which route has the more favorable winds.

Would there not be some benefit in having 2 flights nearby on the same track over uncontrolled space, because they would (assumedly) be able to communicate with each other easily by radio at all times? Or is this not a significant enough factor compared to the rest of it?


text:
What aircraft has the propeller with the highest blade count?
Among all known propeller (unducted) aircraft, which one has the highest number of blades?
Experimental aircraft also count.

The Antonov An-70 is powered by 4 Progress D-27 "propfans", or more technically contra-rotating turboprops, each of which has 14 blades: 8 on the front propeller, 6 on the aft one.
That at 56 blades total puts it ahead of any other modern propeller aircraft. It also has the most blades per engine, at 8+6, while the rest top out at 8.

The Dornier Do-X is an honorable mention at 12 engines with 4-bladed props, also in a contra-rotating configuration, with 48 blades total.
Some Safran's open rotor mockups have shown as many as 12+12 blades per engine, but they're yet to fly, and since 2 engines is the likely maximum for new commercial aircraft, that's still only going to be 48 total.

This raises the question of whether a non-rotating blade counts or not.  These are/will be used to make the airflow behind the moving blades straighter rather than helical.  In straightening the airflow they produce additional thrust so it’s arguable that they are collectively a propeller.

Safrans wacky inflatable turbo prop ?

Do we count turbofans or only unducted props? On a technical level, the blades of a turbofan are just a specific kind of ducted prop and regularly have 20+ blades.

@DarthPseudonym Based on the question, which speaks of propeller aircraft, I would definitely exclude turbofans. But "open rotors" and "propfans" are types of propellers, even if they use other terms to market their more advanced nature. Just like a macbook is still a laptop.

it is limited to unducted only

@Therac thanks for all the data. I want to limit the question to individual props. So that means the Antonov only has 8 blades.

Obviously, things can get a little weird...

@Gabe 8 is the highest any production airplane has anyway. Any more has only been seen in so-far-experimental open rotor designs.

Very interesting that the An-70 has a different number of blades on each propeller.  I suppose the engineering, physics, and aerodynamic testing all played a part in deciding that.

@pr1268 Of course. The two propellers work in very different aero conditions, with high load on the back propeller. To keep the gearbox light, it needs to balance the torque. Also, different blade counts reduce the "chop" sound from blade interaction.

German company MT-Propeller seems to have built and flight-tested 9- and 11-bladed propellers.

(photos from their site)


text:
Does anyone know if RHP (assumed to be rotor horse power) is equivalent to SHP (shaft HP)?
I was recently reading a NAVAER aircraft characteristics summary of the proposed XHCH-1 helicopter of 1951. In the engine section it gives the engine power in "Rhp" and later in a 1956 summary gives the value in Brake HP (Bhp). See https://www.alternatewars.com/SAC/XHCH-1_CS_-_1_November_1951.pdf
Does anyone know if there is a difference between RHP and BHP. I'm "assuming" (and I stress assuming) that RHP is 'rotor horse power' and equivalent to shaft horse power. I'm again "assuming" that the technical description simply changed over the last 70 or so years.
Any thoughts or suggestions from any learned person out there? Thank you, much appreciated.

This answer is somewhat speculative, but I have strong grounds for the speculations, pls read through:
I've only ever seen bhp used as hp measured directly at engine output shaft, although by definition bhp can be measured at any location of the driveline:
Wikipedia: brake horsepower
In the particular case in question, I strongly believe the reason for the use of the obscure "Rhp" is the unusual way the powertransfer is arranged in this contraption: there is/seems to be no mechanical connection between the actual engine and the rotor. The rotor is actuated pneumatically by jets at the tips of the rotor blades.
There would be no point in measuring the brake horsepower or shaft horsepower of the actual engine due to "weird" power delivery.
It is therefore most likely the hp is measured at rotor hub, but since the rotor in this case sort of "is the engine itself", to avoid confusion, Rhp is used to declare power. The declared Rhp probably is shaft horsepower. Kinda confusing though :)
Note: several sources list rhp (with some variation in caps) to mean "rated horsepower, which according to Merriam Webster means "the maximum power an airplane engine can develop without failure when operated continuously or for a specified long period under specified conditions". Even with this definition, I think in this case it refers to hp measured at rotor shaft, produced by the tip jets.

Thank you, I appreciate your help. I suspected the same and yes I should have been more clear, the McDonnell Model 86 XHCH employed pressure jet-tip rotor system similar to the XV-1 and later Model 120 V-1 :Jeep". Thanks again.

You do know the actual mechanism by which helicopters fly? They look so hideous that the earth repels them. Rotors are used for guidance and during landing to force the helo on the ground.

Jet rotor powered by two Allison T56 turboshaft engines that drove a separate forward located compressor stage via a shaft. Compressor stage had bell-mouth inlet. Engine had separate inlet. Pressurised air from separate compressor was ducted to a rotor hub plenum and through rigid blades to tip-jet nozzles at blade ends. Not known if tip-jet burning was to be used, Model 82 (XV-1) did and Model 120 (V-1 Jeep) could. No anti-torque rotor required.

For more info/data on the later (1957) McDonnell Model 120 V1-Jeep see:https://docdro.id/Ncz0Roq

Rotor Horsepower (RHP) is the only way to measure power in a tip-jet propelled rotor system. Power turbines are rated in shaft horsepower, while piston engines are traditionally rated in brake horsepower.
Different measurement methods simply get different names to distinguish how that power level was measured.

Can you point to some documentation to support your answer? I'm not doubting you, but a reference somewhere would turn a good answer into a great one.


text:
Can I calculate fuel consumption for given reduction of drag coefficient?
If plane or car has Cd of 0.35 and 5L/100km, how much will be fuel consumption if Cd is reduced to 0.20 and everything else stay the same(weight, frontal area,power etc)?
Is it possible to calculate this directly if we only know reduction of Cd?

For Airplanes
Yes, use the Breguet range equation. Follow the link and use your numbers: First run it with the L/D that you get with the high drag coefficient, then repeat with the lower drag coefficient. You should look for the increase in landing mass which the drag reduction allows while range stays the same. This increase can be interpreted as the additional payload you can take along. Alternatively, you can look for the reduction in take-off mass to see how the reduced fuel mass due to the drag reduction allows to fly with a lighter plane.
There is more to vary, if you want: Fly both versions of your plane at the same lift coefficient, or go with a lower lift coefficient when drag is lower so you fly both at their optimal L/D. The Breguet range equation shows that range is proportional to L/D.
As an aside: May I assume that you have the zero-lift drag coefficients in mind?
If not, the drag coefficient grows with the square of the lift coefficient and a drag coefficient without an accompanying lift coefficient will be meaningless.
If yes, your values look too high by an order of magnitude.
For Cars
Strictly taken, this question would be off topic here.
Apart from that, the first question is how much of total drag is aerodynamic. Heavy and slow vehicles have non-negligible rolling drag. On the other hand, the drag of a passenger car at Autobahn speed is mostly of aerodynamic origin.
Next question: How much of the energy in the fuel is used to keep the engine and its accessories running? With the AC running at full power, reducing aerodynamic drag will have less effect.
I advise to be more specific for a more precise answer.

Since the question starts with "If plane or car", perhaps it's worth mentioning that the Breguet equation and this answer are only valid for planes, not for cars. +1

@Robe Oops, overlooked that. Thank you for spotting this.

@PeterKämpf 0.35/0.2=1.75 Can I simply divide consumption by 1.75 in car case?

@JurgenM No, only if 100% of the chemical energy in the fuel goes into overcoming aerodynamic drag, which is clearly wrong. Haven't you read my answer?

@PeterKämpf Yes. It cant be proportinaly because resistance forces add up in formula.


text:
Why and how can TCAS and SSR transponder work at the received frequency of each other?
The TCAS sends interrogation pulses at 1030 MHz and receives replies at 1090 MHz. And the transponder replies interrogations at 1030 MHz receiving at 1090 MHz. There could be interference between them, right?

Just like TCAS, the transponder receives interrogations at 1030 and replies at 1090.

"There could be interference between them" -> do you mean a transponder replying to its own TCAS interrogation?

@Bianfable yess, is there a connection between them to inhibit each other while transmiting??

To clarify:

A transponder replies on 1090 MHz to ATC interrogations received on 1030 MHz.

A TCAS elicits nearby transponders replies by sending interrogations on 1030 MHz, and listens for replies on 1090 MHz.

So each one transmits on the reception frequency of the other. This is what allows a TCAS to detect threats, even when there is no TCAS on the other aircraft. As you assumed, both transmitters need to be synchronized to prevent the saturation of the receivers.
For TCAS able to coordinate with the other TCAS, continuous data exchange occurs on 1090 MHz using the mode S transponder datalink, thus this transmission is not performed by the TCAS itself.
Radio equipment global synchronization
There is a dedicated avionics mutual suppression bus/line used to report radio equipment status (transmit/receive), to allow different units to be synchronized.

Source.
Interference prevention specific to TCAS
Several techniques are used, both to limit local radio interferences, and spectrum occupancy.

“whisper/shout mode” used by TCAS on 1030 MHz, to use the lowest possible transmission power to not interfere with aircraft which are too far to be threats (to lessen FRUIT on ground stations)

Beam-steering for top antenna in order to elicit replies only from transponders in a certain azimuth, and adapt the transmission power for this azimuth.

Number of active interrogations limit based on the number of TCAS in the vicinity, to not increase FRUIT beyond limits defined by ICAO equations. This reduces the detection range.

This limit also prevents transponders from being suppressed a too long time. Transponder inhibition occurs after transmission, to reduce the spectrum occupancy. Suppression can also be explicitly triggered by a mode S interrogation from a TCAS.

For a detailed view of the interference limiting procedures, see ICAO Doc 9863)
The next two sections are more about TCAS use of frequencies than a direct answer to your question.
TCAS-to-TCAS coordination on 1090 MHz
In order to coordinate resolution advisories, in particular the vertical rates, both aircraft must be equipped with a TCAS implementing ACAS II protocol (i.e. a TCAS 7.1 or higher), this implies a mode S transponder.
A TCAS sends interrogations on 1030 MHz and listens for replies on 1090 MHz. For this phase the TCAS is relatively independent.
But as soon as a thread is detected, the resolution is coordinated by exchanging data messages between TCAS on 1090 MHz. The TCAS prepares the data to be sent on 1090 MHz, the transponder performs the transmission.
ADS-B Out also uses the transponder datalink the same way.
Overview of ACAS II protocol
We assume the aircraft looking for intruders uses ACAS II.
Observation phase

The TCAS passively listens to transponder replies to ATC (in modes A/C and S) and to DF17 messages (ABS-B broadcast), transmitted by potential intruders on 1090 MHz.

TCAS also elicits direct transponder responses by broadcasting interrogations on 1030 MHz (modes A/C and S). Aircraft in the vicinity reply, and a TCAS can determine each aircraft range and azimuth, and for replies in mode C and S, altitude and vertical rate, identifying possible threats.

Threat detection and resolution

When an intruder enters the closest protection envelope, the TCAS elaborates a resolution advisory, based on its maneuvering possibilities at the moment, and pushes the advisory parameters into register 30 (hexadecimal) of the aircraft SSR transponder.

The transponder sends the resolution advisory in a DF16 message (on 1090 MHz), this message is received by the TCAS of the intruder, and, assuming the intruder has an operational TCAS, is used to  determines a complementary maneuver.

If the intruder TCAS implements only ACAS I, the process cannot be coordinated and stops here. Else a coordinated TCAS-to-TCAS exchange continues on 1090 MHz via the transponders. The resolution advisory may be adjusted by each TCAS, until the threat is cleared.

More

ACAS protocol in ICAO Annex 10, vol 4.
Mode S in ICAO Doc 9688.
Mode S messages in The 1090 Megahertz Riddle.
My answer to Why is a Mode S transponder required to be installed and working for TCAS II to be operational?


text:
Intersecting Boundary Layers in XFLR5
I am analyzing an airfoil called RevE-HC. While doing a type-1 analysis, I encountered that the BL is intersecting towards the trailing edge. The panels are refined and the geometry is also smooth. Any clue as to why this is happening and how to resolve this? Similarly, the Cp vs x graph shows a double peak towards the LE and a sharp peak at TE. What does that mean? Any help would be appreciated.
The data is for zero AoA.
The data:


text:
Does an airplane fly less or more efficiently after a mid-flight engine failure?
When a multi-engine airplane loses an engine mid-flight, does it consume more or less fuel per distance travelled during the remainder of the flight with fewer engines operative? In other words, is the maximum range extended or reduced?
And for what reason (if that's an answerable question)?
As far as cars are concerned, 4-wheel-drive is less efficient than front-wheel-drive, see:

https://engineering.stackexchange.com/questions/26704/is-it-true-that-four-wheel-drive-is-less-efficient-and-if-so-why
https://mechanics.stackexchange.com/questions/48012/where-is-extra-fuel-consumption-in-rwd-and-awd-cars-dissipated

I am basically asking the same question for airplanes.
In case it matters, please distinguish between aircraft engine types in your answer.

This is probably a duplicete question, here is some numbers for A350:
https://aviation.stackexchange.com/questions/104539/cruise-with-a-single-engine-in-a-twin-good-idea/104591#104591

In almost all cases, the maximum range is reduced.
This is because the in-operative engine causes substantial drag.  In addition, it causes an asymmetric flight condition that requires deflection of control surfaces and flying at an unusual attitude -- all which cause more drag than nominal cruise flight.
It is not entirely an apples-to-apples comparison because the damaged aircraft will likely also reduce its flight speed.  Drag will be reduced, but flight time will also be increased.  Most commercial aircraft strive to fly at slightly faster than the best range speed.  Fuel efficiency (pounds per mile) is very important, but we'll take a small penalty because we also value our time (get there fast).  A common rule is to fly beyond the best range speed at a speed where your fuel efficiency is 99% of the best efficiency.
So, this means that an aircraft in an emergency could get a very small efficiency improvement by slowing down slightly.  Not enough to overcome the added drag.
There were some propeller aircraft that could increase their range by shutting down one or more engines.  They could feather their propeller to minimize the drag of the non-operating engine (jet engines don't have this option).  And then they could slow down as well.
There are also some three-engine helicopters (think Coast Guard Rescue) that shut down one engine in cruise and then start it back up when they get to the rescue site.
In cruise, with three engines operating, they operate at a relatively low throttle setting -- where they aren't as efficient as they could be.  So, by shutting one down, the other two can operate at a more efficient point.  This is particularly true when they're flying at a best endurance point vs. a best range point (this is slower and lower throttle).
In cruise, if one engine were to fail, you have time, power, and margin to get the third engine fired up.  As you get to the rescue site (or when you find the rescue site), you start up the third engine to maximize redundancy during the rescue hover -- where you have no time or altitude for a failure.

"They could feather their propeller to minimize the drag of the non-operating engine (jet engines don't have this option)" You can't feather the fans of a variable pitch turbofan?

You can feather the blades of a variable pitch turboprop -- and it would be interesting to look at flight models for 4-engine turboprop aircraft (C-130, P-3, etc) to see if there were cases where that was advantageous.  Some turbofans have variable guide vanes - which (ignoring variable area nozzles) is the only type of variable geometry I am aware of on turbofan engines.  VGV presents a limited amount of control and is insufficient to stop rotation and dramatically reduce windmill drag.

Also, your typical jet will not be able to fly as high and thus suffer more drag.

Interesting. I would have assumed that, since air resistance should go with the square of the speed (correct me if that is not true), lower speed could more than compensate for higher drag. (That is, the emergency range -- optimizing altitude and speed -- with one engine out should be higher than the regular range with all engines.) Not true?

@Peter-ReinstateMonica If that is the case, then there should be cases where flights would travel at lower speeds and altitudes without there being an emergency - i'm thinking cases where speed is unimportant and fuel savings preferred. does such a thing ever occur?

@Peter-ReinstateMonica There are two major contributors to drag -- the parasite drag, which increases with speed is squared like you're thinking.  And the induced drag, which actually decreases with speed (it is proportional to 1/V^2).  Consequently, there is a strong tradeoff resulting in a minimum drag airspeed.  This is unlike cars, trains, and boats -- none of which have to 'pay' to hold themselves up.

@Michael  For aircraft, there is always a speed for best range.  Some will choose to fly faster than that (and therefore could slow down to save fuel), but typical commercial operations are at very close to the optimum range flight conditions.

@RobMcDonald Yes but that's at optimum cruising altitude, right?

There will be an optimum speed for any altitude.  It won't be as good as the best speed and altitude combination.  For turbofan transport aircraft, the best operating point (for range) is very close to M0.85 at 36000ft.  Slowing down will make your range worse. Look at the (SR) chart I attached to this answer https://aviation.stackexchange.com/a/106528/65514 .  Adding one engine out drag will only make things worse.

Fuel consumption for OEI is significantly higher per mile.
Refer to this question for an example of fuel flow: Does one-engine cruise consume less fuel than two-engine cruise?
There's an 8% reduction in fuel flow per hour, but a 16% loss in airspeed, and that's assuming the same altitude. The altitude for single engine cruise will be lower, which makes matters worse.
It's in part the extra fuel reserve required for OEI operation that determines ETOPS range limits.

In a vehicle with 4WD, you have one engine driving all the wheels. This drivetrain contains gears, universal joints etc. that have friction. In this case, disconnecting drive to two of the wheels reduces friction which reduces fuel consumption.
In an airplane, each engine is (generally) a self-contained unit. You don't have one engine driving multiple propellers.
You do reduce friction losses in the engine itself when you shut down an engine, but those savings tend to be smaller than the extra drag you produce by having an unpowered propeller or fan in the airstream.
There are exceptions:

the HS Nimrod (a 4-engined maritime patrol aircraft) could be operated on two engines. The aircraft would be flown at low speeds during the patrol, and at those speeds it was more efficient to have two engines running at full throttle, instead of running four engines near idle.

The Fairey Gannet used the same principle. It had a Double Mamba turboshaft (i.e. two Mamba engines in a single casing) driving contra-rotating propellers. One of the engines could be shut down. Again, this was used for maritime patrol where it had to operate at low speeds for long periods.


text:
How does a flex takeoff increase fuel consumption compared to a TOGA takeoff?
I read this also in the FCTM green operating procedures that doing a flex takeoff increases fuel consumption as we have a longer takeoff phase

When you ask your instructor about this you may share the answer here :)

I’d change the question to something more along the lines of “how does assumed temperature takeoff affect fuel consumption.” Flex is not the most common term for the concept. The question also asserts that reduced thrust takeoff causes an increase in fuel burn without reference or establishing that as a fact.

I would like to run numbers to be sure (don’t have time at the moment), but I believe in general it’s because flex takeoff doesn’t make most efficient use of fuel when viewed for the whole flight. Assuming standard day, no wind, jet transport aircraft trip fuel is minimised when the optimum cruise altitude (maximum specific range) is obtained as quickly as possible after takeoff. Choosing to use less than maximum thrust will make the aircraft stay lower (and in dirtier configuration) for longer, thus increasing fuel consumption over the theoretical optimum. Comment instead answer, as no proof.

Fuel consumption for the flight will be slightly higher when carrying out a reduced thrust takeoff.
Simply put, the flight is operated at lower, less efficient altitudes for a tiny fraction longer, so less time is spent at more efficient altitudes. It’s a rather small amount, but nonetheless measurable.
From Airbus‘ Getting to Grips with Fuel Economy:

Compared to a full thrust take-off, flex thrust will generally increase fuel burn. The increased time at low level offsets the slight reduction in fuel flow induced by the lower thrust.

The source goes on to list typical fuel burn increases, e.g. for an A320 it’s 1kg (!) when taking off in CONF 1+F and 5kg in CONF 2.

I am afraid it is just the opposite, even though the take off phase is longer the fuel consumption is reduced.
The slope of the delivered power to fuel consumption goes decreasing well before reaching full thrust.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What's the reason for this frog finger shaped parking lots at Köln/Bonn airport (CGN)?
Most aprons of international airports are vast areas paved with asphalt or concrete. However, I noticed this structures in the south of Köln/Bonn airport (CGN) (Google maps link):

Does anyone know what's the reason behind this unusual shapes?

EDIT:
Just found the Royal Airforce Base Brüggen with lots of dispersal areas. Some with hangars, some with banks of earth like in the black/white picture in aeroalias' answer. It's also a perfect example for the conceptional layout. (And between the areas in the south, there are even typical bunkers for explosives)

Ex-military airfield.  They are "dispersal" areas.

Picking nits: Frogs don't have "fingers" :)

@CGCampbell: Say who?

@Simon is right. The 'parking lots' are actually dispersal areas in the (previous) military airfield at that place.
The Cologne Bonn Airport started out as an airfield for Luftwaffe in 1938 in what was a former artillery shooting range. After war, RAF took over the airfield, with the first civilian flights in September 1950. The airfield was converted for full civilian use only in 1957.

"Koeln-Bonn-Airport14" by Borsi112 - Own work. Licensed under CC BY-SA 3.0 via Commons.
The original 1886m (32L/14R) runway is near the bottom of the image (near the dispersal area). The top runway, 3800m in length (32R/14L), was opened in 1961.
In military sirfileds, the parked aircrafts are usually dispersed in order to reduce vulnerability in case of an air raid. According to US Army Field Manual 5-430-00-2 (accessed via globalsecurity.org)

If dispersal of aircraft is possible and consistent with active defense measures, varied parking patterns provide fewer lucrative targets for indirect-fire weapons.

It also gives some (conceptual) layouts for aircraft dispersal:

Source: globalsecurity.org
The aircraft are parked in such a way as to reduce the probability that more than one aircraft is hit simultaneously. These dispersal areas have banks of earth or small walls around them to reduce the aircrafts’ susceptibility to the effects of bombs landing nearby and to reduce collateral damage in case one of them blow up. It might've looked like this:

Source: portalnews.gq

Thanks for the informative answer. I had trouble to really understand what "dispersal" meas, but now I understand it's to reduce damage during air raids.

I have to comment that this is an awesome and informative answer and you've properly attributed the images... well done

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What aircraft is this?
Yesterday, I saw a strangely-shaped aircraft in the sky.
Searching for "aircraft silhouette" on the web, I found this picture within seconds on a vector graphics site. It describes perfectly what I saw, so there is nothing else I can say about the aircraft.

But I did not find any information what aircraft it is.
The wings are not swept and those stubs at their front look like propeller engines. But it seems as if there are fans at the rear of the fuselage...
So, does anybody know what aircraft this is?

EDIT:
@DeltaLima: Not bad. I'm pretty sure that this picture really shows an A-10. The picture is part of a larger one, and the other aircrafts were military ones, too.
I saw the aircraft it in Geneva, near GVA airport. Now, I doubt that I saw an A-10 there, as the last A-10 left Germany a while ago, and afaik there are none in Switzerland or France. If I neglect the stubs (which I may have noticed on the picture only), it could also have been  one of those at the GVA airfield, which have a comparable shape.

EDIT 2:
Meanwhile, I'm quite sure that it was one of those labelled A, B, C on this picture from google maps (though, B and C are pretty much the same):

The airfoils as well as the elevator were definitively not (significantly) swept and there was no gap between the airfoils and the fans in the silhouette.
Maybe, it was one of those, maybe, it was a similar aircraft. May be, it really was an A-10.
So, this are three 'maybes'. Lets add another:
Maybe, we put this question on hold as there will be no definitive answer.
But thanks a lot for your contributions!

While Switzerland and France don't have them, they might have US planes en route to middle east. Unless they were low as if on departure or approach; they have decent ceiling and would have no reason to cruise low over Europe (and, as all US Air Force aircraft, they are aerial refuelling capable, so they wouldn't really need to make stops; though they could, of course).

Hi sweber, I've voted to put this question on hold because I'm concerned that it's no longer "What is the aircraft that this silhouette represents" (which is a good, answerable question), and start being "What airplane am I thinking of?" (which is likely to turn into a guessing game)

@JanHudec - it is not true that all US Air Force aircraft are air-refuelable, though most certainly are (and the A10 definitely is)

@SSumner: True, some helicopters are not and some trainers/utility/transport are not either. But all fixed-wing combat aircraft should be.

@JanHudec - I'd point out that "all air force aircraft" and "all fixed-wing combat aircraft" are two very different things...not close in size at all. Though we're just arguing about semantics here.

The three aircraft you label differ from your silhouette significantly. They all have forward-swept wing trailing edges and tapered tailplanes, and none has protrusions forward of the wing leading edge, which the silhouette has.

It seems to me that it is an  Fairchild Republic A-10 Thunderbolt II, commonly know as "Warthog".

(Source: wikipedia)
The "stubs at the front of the wings" are pods in which the main landing gear is stowed after retraction.

If you are in doubt that it was military, it might have been a business jet like the Cessna Citation II / Bravo / Mustang, which has a similar planform albeit without the stubs in front of the wing.

Using the replay function of Flightradar 24 you may be able to find out which exact aircraft it was. The coverage at low levels is not so good for all aircraft though.

Not bad! However, I'm not that sure anymore that I saw an A-10. See my edit.

Yes, I meanwhile think it was one like this business jet.


text:
What do the components of a shockwave's visual artefacts represent?
On a recent flight, I was able to observe visual artefacts over the wing (Boeing 767-300 in cruise).
You can see them in the image below; they look like creases in a surface. They moved forwards and backwards over the wing. Later in the flight they extended to the end of the wing; at this point, they extended no further than a few metres.
I've added red and yellow arrows pointing to their ends - as you can see, one of them appears to fork.

Adjusting brightness and contrast makes the artefacts more visible:

I understand that this artefact is caused by a shockwave produced by supersonic airflow over the forward part of the wing.
I'd be interested to know: what exactly is it that I am seeing?

Is the artefact projected onto the surface of the wing, like a
shadow?
Is the artefact a refractive distortion caused by different air
densities?
In this image, there appear to be two creases, one behind the other -
what do they represent?
What does the fork at the right-hand end of one of the 'creases'
represent?
What do the darker and lighter portions of the 'creases' represent?

You see a mirage, that is an image deformed by the change of refractive index, due to the huge change in air pressure / temperature / density. See shadowgraph. As the index changes, the phase of the light is altered, creating interferometric shadows. You can compare this with the effect visible over a hot road surface.

I'm not sure this question is about aviation, also given @mins' comment above.

Related (not a duplicate): Is the visual artifact seen over the wing from different air densities, a reflection or something else?

This phenomenon is actually often used to visualize shocks in wind tunnels and experiments, see Schlieren photography
And since we have an 'aerodynamics' tag, I don't think this is off-topic.

@mins It's not clear from your answer whether what I am seeing is the wing, with a narrow line of distortion in the image caused by the change in refractive index, or a shadowgram of that refractive distortion projected onto the wing by the (nearly) overhead sun.

The wing is illuminated 1/ By the scattered ambient light; 2/ By the ambient light altered when crossing the shockwave refractive volume:  Refraction creates a shadow more or less visible by diverting a part of the light elsewhere; 3/ The direct light from the Sun (if any) which creates another (stronger) shadow. Then you see this illuminated wing through the shockwave. The rays are again curved / folded by the index changes, so the portion of the wing may not match exactly the surrounding details (that's not visible on the image)...

... About shadow creation: Can air make shadows? --- and Why do the edges of glass blocks create a shadow?. See also Schlieren imaging pointed by @ROIMaison (video)

What you see are changes in the refractive index of the air, caused by the different pressure and temperature at different places. Sometimes, with the right conditions, there may be zones of transient condensation of water vapor, that add to the effect...

In this video, the shock wave produced by a big explosion is clearly recorded by high-speed cameras... http://www.wimp.com/invisible-worlds-explosion/


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What plane is most similar to the LEGO Firefighter Aircraft 42152?
I want to identify a LEGO Firefighter Aircraft: LEGO set #42152
Is this model based on a real plane? Does anyone have any ideas?

I don't think this is a replica of an actual model of aircraft, however I believe that this set has taken a lot of inspiration from the Canadair CL-415 both in the design and more so in the color scheme.

The only other real example of a twin prop aircraft used for firefighting is the Antonov AN-32 and that doesn't share any more characteristics with the set than the Canadair, and certainly doesn'tt match the paint scheme.

While it could be any of the previous aircraft from the other answers, I would suggest that it is a DHC Dash 8 water bomber. Both are twin turboprops with a t-tail configuration.

This is supported by the Lego model's main landing gears being attached to the engine pods, which is a characteristic of the Dash 8. The CL-415 MLG is fuselage-mounted.

Yeah, I think the Dash-8 is a good shout, I didn't even think of that one.

That's for sure a nice mixture inspired from the turboprop-powered airplanes by De Havilland Aircraft of Canada (DHC).
In particular, the airframe looks like the one of a DHC-8 aka Dash 8: T-tail, landing gears retracting in the nacelles, tapered wingtips and general proportions they all point in that direction:

The paint job is anyway unmistakably the one of their firefighting flying boats, like the CL-215 or -415

The flat red belly seems to have also been taken from this latter.

(Picture source)
I suppose they've kept the set unlicensed in order to leave the price as low as possible. They did a nice mix, have fun with that set

All other pictures from Wikipedia

You can make the propeller spin with a lever in the belly?! That is too cool.

Near as I know, the aircraft featured in that technic kit may be a generic looking “water bomber” that is not an exact match for an actual aircraft currently in service. From the color scheme and lines of the aircraft, it is very similar to a Viking Aircraft DHC 515, which is based upon DeHavilland/Canadair’s CL215/415 series of amphibious aerial firefighting aircraft.

The DHC 515 is a large aircraft, nearly the size of a 737, specifically designed for loitering over wilderness terrain near hotspots, then can make deliveries of up to 1600 gallons of water on a fire with or without chemical retardant added.  It can then go scoop another 1600 gallons of water directly out of a nearby lake and return to the loiter area. This gives the aircraft probably the greatest versatility for fighting wildfires out of any airframe out there.
The Lego Technic model appears to match the DHC 515 in terms of the shape of the nose as well as it is a high wing design but does not appear to feature the ‘flying boat’ fuselage superstructure and the outrigger floats nor the cruciform design of the empennage that the Viking DHC 515 features.

Nearly the size of a 737 Not quite. Yes on wingspan. But comparing to 737 Classic, MTOW for 737 is about 3x, and length about 1.5x the DHC515.

Dimension wise, that’s pretty comparable to a 737 Classic. The slugs were heavier, mind you, but the DeHavilland ‘Ducks’ aren’t light either!

The 737 is much longer - about 100 ft even for the Classic, versus 66 ft for the DHC-515. The Boeing's cabin is also 4 times the size. They're only similar in wingspan, and only because the Boeing's wing is swept. Next to one another, it would be obvious one's much larger. Now, the Beriev Be-200, that's a 737-sized amphibian.

I think the phraseology was nearly the size of a 737. At any rate, it doesn’t matter.


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Is there an equation to solve for bank angle using load factor and wing loading as variables? If so, what is it?
The problem is that I'm looking for both wing loading and load factor as variables to solve for bank angle and cannot find any equation that represents this. After searching various websites and not coming across the right equation, I asked an A.I. chatbot. It gave me the following equation:
$$ \theta ≈ \arctan \frac{1}{N⋅\frac{W}{S}} $$
$\theta$ = Bank Angle
$N$ = Load Factor
$\frac{W}{S}$ = Wing load, Total Weight of Aircraft ($W$) divided by wing area ($S$),
However, when the load factor input increases, the denominator increases, making the bank angle smaller. Wouldn't a higher load factor equate to a greater bank angle? What would the correct equation be?

I don't think your formula is correct, see this question

"I asked an A.I. chatbot..."  Well, there's the first problem. The current state of AI "chat" is an answer that sounds right, and actual veracity is an occasional byproduct of that, nothing more.

Does this answer your question? How can I compute the minimum possible bank angle for a turn with a given radius?

Consider that load factor is a multiplier for the weight of the aircraft, and as such it doesn't care what that weight happens to be. A heavy and a light aircraft flying the same path (i.e. maintaining the same bank angle and speed) will have the same load factor, regardless of weight (or wing area).

Wing loading would have nothing to do with it.  You can put it as a term in your equation if you want, as long as you make sure to multiply be zero!  Now if if you are trying to solve for turn radius, and airspeed isn't a variable but lift coefficient is -- then the wing loading would make a difference.

Ok thank you! I just found this website that talks about wing loading online, https://howthingsfly.si.edu/ask-an-explainer/what-wing-loading, and I wasn't sure if it directly correlated with maximum bank angle under a given load factor. Do you know what the equation is to solve for bank angle under a given maximum load factor? Also since wing loading is correlated to lift coefficient, and lift coefficient is correlated to maximum turn radius, wouldn't wing loading be considered a variable in maximum bank angle?


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Is the F-117 actually inferior in its RCS than the F-22 and B-2?
This might seem like a purely opinionated thing, but bear with me. I think there is some general statement to be made here not based on guesswork.
Question: Is the F-117 actually inferior in its RCS (Radar Cross Section) to the F-22 and B-2?
My own attempt at arriving at an answer:
The F-117, the F-22, and the B-2 don't have any publicly available information on the actual RCS to my knowledge, nor do they offer any details disclosed about the exact makeup of their RAM (Radar Absorbing Material). So, how can anyone actually know this?
I watch a lot of aviation documentaries, read sometimes the odd article, on Wikipedia, etc. -- it seems most sources simply make the statement that, the F-22 has a "much better RCS" than the F-117. Same with the B-2.

What do I mean by superior or inferior? Well, therein is the answer to the question as well.
I think the answer is: It depends.
The RCS depends on a number of attributes, certainly at least: The orientation of the object being measured, the frequency of the radar system, the weather.
Even in this simple 2D example, for a non-stealthy aircraft, its RCS profile is non-uniform.

For an aircraft with flat, angular surfaces like the F-117, disregarding possible differences in RAM (Radar Absorbent Material), an angled surface will always reflect away a signal hitting it at an angle, better than a curved surface. In a simplified setting, an angular surface, except when it is hit directly perpendicular with a signal, should reflect away all the energy.
In reality, of course, the asperity of the surface of the material will ensure that some energy does get reflected. However, it is probably fair to say that you'd prefer an angled surface hit at an angle over a curved surface being hit with a signal, given that everything else is equal and remains unchanged.
Basically, this appears to me to imply that (again disregarding RAM differences) the F-117 actually enjoys superior RCS from any angle except the set of angles where the aircraft happens to expose one of its angled surfaces perpendicularly to an observer.
How big is this set? Probably quite big. However, this is something that can be planned for and mitigated. E.g. when facing adversaries on the ground, the F-117 is observed from below, and from below, the F-117 has only one surface. The F-22, has a lot of complex geometry, in comparison.

So roughly speaking, unless the F-117 flies directly overhead an observer, why would the F-117 have inferior RCS?
As for the RAM, it is again sometimes said that the F-22 enjoys superior RAM -- but again, this is not publicly available, is it? It is merely an assumption, that the F-22 has superior RAM. The exact makeup and details of the RAM for all of the abovementioned aircraft remain classified. Correct me if I'm wrong.

Why wouldn't newer aircraft have better 'stealth'? A couple of decades of development would bring better 'performance'. It has been thus with all tech. Cars, computers, aircraft.

It's not that simple. Remember radar waves are interfering waves, so that the reflections from two surfaces could interfere can create a third path of strong echo. With 1970's computing power there's essentially no way to know this.

Other than the (entirely reasonable) "decades newer is likely to be better" conjectures, it's unlikely that there is enough authoritative information available in the public domain to provide an actual, objective answer to this.  Seems like this is unanswerable except with opinions & conjecture.  Some of that conjecture may be intuitive, reasonable, and believable (i.e. the comments above), but the question itself is a question about speculations, and probably a poor fit for Av.SE.

@RalphJ Wouldn't that make the answer quite simple, in the same vein as I've been suggesting might be the answer: This is not publicly known, and any documentary or article that attempts to claim otherwise, are simply, wrong? This is not public information, and therefore, we simply don't know. The F-117 might have better RCS from some angles, or it might not, depending on the RAM and other attributes?

If you'd like to ask a question and self-answer, that's completely acceptable. However, you should ask the question in the "Your Question" box, then provide the answer in the "Your Answer" box. Putting the answer in the question isn't how things work here.

The few people that know the lab measurements, aren't at liberty to tell. So the only public information on stealth comes from modeling. However, stealth is complicated. It's not just specular reflections, which are easily modeled by physical optics, but a number of others:
.
As a result, the outcomes of modeling are all over the map. RAM is prominently excluded from most of these models, and some even miss radar blockers.
But combat experience allows us to make an estimate. It's well known that one F-117 has been shot down by a rather outdated radar-guided missile.
Some early media reports of the shootdown have speculated that the missile was "eyeballed", or that bomb bay doors were open. The former is impossible: the crew's located inside a windowless room, and the missile isn't on a turret. Open bay doors would explain detection, but not a shootdown, since these older SAM have separate, mechanically steered search and targeting radars, taking about a minute to launch. There's a Hungarian-written SAM simulator which shows the controls and operation of the S-125.
The reports available today show the SAM was operated in radar mode, as intended, tracked / lost track of the target 4 times, and ultimately one of the two launched missiles achieved lock and hit; the other missed. Here is the report from the battery's commander, translated from Serbian:

I arrived at the firing position around 20:30. There were no nearby targets in the air, but some were at greater distances in various azimuths. Suddenly, on the observation radar display, at an azimuth of 195, I spotted a target at a distance of 23 kilometers.
At a distance of 14-15 km and an azimuth of 210 degrees, the firing officer, Lieutenant Colonel Zoltan Dani, ordered to search for the target. The targeting radar's radiation was turned on. We radiated for more than 10 seconds unsuccessfully.

I saw the target again at an azimuth of 240 degrees and a distance of 14 km. The guidance officer's dials clicked, but the operators lost it. Just when I thought this attempt would also fail ... The dials clicked, and the operators locked on. Stable tracking, azimuth 242 degrees, distance 14.5 km.
The first missile launched, then the second after 5 seconds. Muminovic reports the first launched and locked, the second launched but not locked. (20:55, end of engagement)

This report is consistent with how the SAM operates. It shows that the S-125 was able to detect an approaching F-117A at 23 km and engage it at 14.5 km. That was its very limit, as indicated by spotty tracking. The aircraft was detected and lost again 3 times, until the final lock on and shot. The exact missile used was "5В27Д" aka "В-601Д". Its specifications are as follows: minimum target RCS - 0.3 m2, maximum range - 28 km.
From this, we can estimate that, even if the missile's "minimum RCS" is specified at maximum range, the F-117's RCS, with RAM, can't be lower than (14.5/28)^4*0.3=0.02 m2, and is likely closer to 0.3 m2. Or, in dBsm terms, between -16 and -3 dBsm. The former figure is optimistic, since minimum RCS is normally specified at the best-Pk range, which is usually around half of maximum range.
While the claims about -40 dBsm for the F-22 and -30 for the F-35 should be taken with caution (it's direct head-on at one frequency only), the consensus from most modeling is that it's no more than -16 dB without RAM, and likely -25 dBsm with RAM for the F-35.
The only other stealth aircraft with RCS without RAM in the -3 to -16 dB range is the Su-57, modeled here, which is consistent with the manufacturer's patent. And it's a sure bet that Lockheed's 3rd and 4th attempts at stealth are better at it than Sukhoi's first, especially being more of a "Silent Flanker" than a pure stealth design.
What has improved in technology since the F-117A includes:

Composite skins. The F-117A is all-metal, and metal panels are susceptible to creep deformation. CRFP panels retain their initial shape much more accurately.
Composite sandwich panels. While the main skin has to be a radar reflector to shield the internals, semi-transparent sandwich panels are very useful for control surfaces, which need to move.
S-ducts that are heavy, but very effective at concealing the engines. The F-117 uses radar blocker grates to shield them, visible on the outside. Out of modern stealth aircraft, only the Su-57 uses that design.
Thrust vectoring, allowing for quiet maneuvering without using the control surfaces, which make the shape less-stealthy while in use.
Edge treatment. Semi-transparent fiberglass composites are used on the edges of modern stealth aircraft, greatly reducing traveling waves and edge diffraction.
Data from the F-117A itself - lessons about what works for stealth and what doesn't, in a full-scale aircraft going through a brief but real lifecycle.

More recent reports also indicate that another F-117 has been detected and shot at by a similar SAM over Serbia, but only received minor damage. This article, coming from that plane's pilot, also details the lengths the US went to in order to keep the F-117s safe: avoiding "double-digit" SAM (SA-10 and above), providing F-16CJ SEAD escort with HARMs, and EF-111 and EA-6B providing radar-jamming.
Protecting a stealth aircraft with so many non-stealth ones isn't a sign of great confidence. But flying it in combat conditions was definitely worth it, for valuable lessons to improve future stealth designs.
Regarding composite vs metal, this is what the "flat" skin of the F-117A actually looks like. The wrinkles around the metal are pretty good at reflecting radar. This is further evidence that the F-117A is between 0.1 m2 and 1 m2 in terms of RCS. 0.3 m2 seems most likely, since that's what the S-125 is advertised at.
That said, the F-117A does have some great stealth features that have been lost in newer aircraft. One is the narrow slotted exhausts - they provide the lowest infrared and sound signatures of all shapes. Another is the first IRST on a US jet, which unlike even LPI radar is completely immune to interception.

Conclusion: the F-117 was an early attempt at stealth, made with limited knowledge, tools, and materials. Its shape reduces specular reflections, but these are just one of many kinds. It's likely that its designers have under-accounted for many variables.
Combat experience indicates the F-117's actual RCS to be significantly higher than predicted by a basic physical optics model. In particular, its S-band RCS is likely on the order of -3 dBsm, at best -10. Modern stealth design is considerably better in that regard, even for the Su-57, the least-stealthy of 5th gen fighters.

Very nice answer. However, what are your thoughts on angular surfaces I mentioned briefly in the question? Do you believe that the F-117A enjoys lower RCS viewed from certain angles, such as the underside (which appears like a flat surface) at angles that are significantly offset from the perpendicular?

@AlphaCentauri The flat surfaces are excellent at minimizing specular reflections. But it appears that returns at the edges weren't as well-addressed in the F-117A. An aircraft's RCS is a 4D function, so there likely are angles from which the F-35 would be a larger target - in particular, from the rear. But from the bottom, the F-117A was a large enough target for three different radars (search, targeting and missile's own) to find and track it in Serbia.

Nice story bro but you should mention that F117 bomb bays were open or bring arguments against it as several sources report this version of the facts.

@HAL9000 Some of the older reports have speculated on this version. But it's not consistent with the Serbian report, or with the report from another USAF pilot, who said the mission lacked EA-6 support.

A lot of the answers above sound about right with the lack of public data and the f117 being older should have an inferior RCS. After reading both Skunk Works and Stealth, the info shows those answers to be correct. One thing mentioned in the books were round shapes actually have a better RCS than flat shapes placed properly. This is why the competitive B2 is still in wider use (of course there’s other reasons too). This is also a reason why the F22 has its more rounded shapes. Round vs flat comes down to radar reflection and sparkling. Another big reason is the windshield on the f117 is flat and they went though some massive struggles on its coatings/materials (that constantly deteriorate) to make it low radar and actually transparent for the pilot to see out of. This does not even mention the other jamming and false signals newer aircraft utilize to mess with radar. Hopefully this added some more insight vs the “classified and outdated” answers even though they are perfectly right.


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C-130 supreme propulsion on thought experiments
I watched some youtube shorts of someone playing a simulator and having planes start off from the roofs of tall buildings and trying to takeoff before they crash. For some reason, that creator mentions that the C-130 is the "king" of these challenges. Indeed, assuming he handles them with equal skill, the C130 seems to outperform even fighter jets on some of these cases.
Why is that? Does it have to do with propeller usage?
Video showing the challenge:
https://www.youtu.be/shorts/3QNKdHzGPlw

What exactly is the challenge?  A very short runway, then a cliff, then achieve level flight without hitting the ground at the bottom of the cliff?  How big?

@CamilleGoudeseune The "challenge" is to take off from the helipad of the Telekom Tower in Kuala Lumpur.  This, I believe, is the video in question.

@Tevildo quite right, I was just not sure if the rules allow the posting of links, so that's why I didn't mention it

A rocket based airplane would be the actual king of this challenge. The Soviet have tried to launch Mig-21 from rocket racks.

There are competitions for short takeoff and landing. If I am to believe the Youtube videos titles, 9'5" for landing and 6' for take-off, but I didn't check out what the headwind was.  In other words, instead of going off of Telekom Tower, you could probably do it off your home's roof ;-)  https://nationalstol.com/

Presumably that 'creator' doesn't know about the Fiesler Storch, or the Westland Lysander, etc. etc. Not to mention, Natter; Harrier; Osprey and the like.

Fundamentally, the C-130 has a very low stall speed.  Mostly because it has a very large wing for the mass of the aircraft.  The C-130 is not meant to fly fast, it is meant to fly slow allowing it to operate from unimproved landing sites.
Fighters have a very high stall speed.  They are meant to fly very fast.
The takeoff process is all about acceleration.  How quickly can you accelerate from zero airspeed to above the stall airspeed.  If the stall airspeed is very low, then you have a great advantage.
Propellers are also very good at static thrust.  They can have very good acceleration at brake release.

It may also help that the sim version of the C130 is probably modeled as as flying with no payload. For truly absurd C130 STOL see https://en.wikipedia.org/wiki/Operation_Credible_Sport#Development

@GremlinWranger, Thanks for making me aware of this, what an amazing project

@ROIMaison, decided to turn my comment into an answer

Probably the first point is to be very cautious saying much about real aircraft from observing a simulator, especially outside the normal flight regime where the flight model may be less accurate and physics of things like rolling off an edge fairly benign.
Second as someone who made their own fun with earlier versions PC flight sims I'd assume anybody doing videos of absurd take off and landings would be at least setting a favorable wind conditions if not getting silly. An empty (simulated) business jet can get in and out of all sorts of fun places if given a perfectly steady 80 knot head wind.
In real world terms this sort of shenanigans is the domain of STOL of which the C-130 is an example though by no means 'supreme' at it. Some purpose built designs have normal take off runs that would allow operations across a suitably flat rooftop without even needing a drop off the end. Examples of some real world operations here. See also entire domain of Bush Flying.
In answering the headline question, the supreme propulsion for short take off is rockets, see Operation Credible Sport which attempted to use substantial amounts of rocket thrust to get a C130 into a sports field and then out again fully loaded, and suitably modded into the sim would make LANDING on that rooftop feasible. We also have various rocket launched fighters, that would happily take off from a rooftop and keep going up.

For those who consume janky 10 year old gaming videos, youtube of a not very successful attempt to model and fly credible sport in the game/construction kit Kerbal Space Program.

I believe you will be pleased to see that the modeled scenario is based on real life. Luke Czepiela did take off from and land on Burj Al Arab's helipad with a Carbon Cub specifically designed for him.

Yeah, especially wind around buildings can be very tricky as explained in length in the Red Bull landing video on this strange building's helipad.